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3 years ago

What information can be deduced by analyzing the graph?

Physics

2 answers:

snow_tiger [21]3 years ago

5 0

The answer is D, Wavelength and frequency have an inverse relationship.

erma4kov [3.2K]3 years ago

4 0

your answer is - the wavelength and frequency have an inverse relationship, so your answer D.

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1- Show that the equation d=vt2<br> dimensionally correct or not?

TiliK225 [7]

- Show that the equation d=vt2 dimensionally correct or not? ... nalsn7 is waiting for your help. Add your answer and earn points.

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2 years ago

If the work done is 360 J, and the force used to do the work is applied over a distance of 12 m, what is the value of the force

hoa [83]

The formula for work is $W=Fdcos\theta$ . Plugging in the numbers, you get:
$360=F(12)cos(0)$
$F= \frac{360}{12}$
The answer is 30 N.

3 0

3 years ago

Read 2 more answers

The Hulk lifts a train into the air. He applies a force of 4000 N and lifts the train 10 m. How much work does the Hulk do on th

Sphinxa [80]

Answer:

hulk smash

Explanation:

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3 years ago

stion 13 of 20 : Select the best answer for the question. 13. If a certain mass of mercury has a volume of 0.002 m3 at a tempera

Free_Kalibri [48]

Answer : The correct option is, (A) $0.002010812m^3$

Solution : Given,

Volume of mercury at temperature $20^oC$ = $0.002m^3$

As we know that the mercury is a liquid substance. So, we have to apply the volume of expansion of the liquid.

Formula used for the volume expansion of liquid :

$V_{T}=v_{1}[1+\gamma (T_{2}-T_{1})]$

or,

$V_{2}=V_{1}[1+\gamma (T_{2}-T_{1})]$

where,

$V_{T}$ = volume of liquid at temperature $T^oC$

$V_{1}$ = volume of liquid at temperature $20^oC$

$V_{2}$ = volume of liquid at temperature $50^oC$

$\gamma$ = volume expansion coefficient of mercury at $20^oC$ is 0.00018 per centigrade (Standard value)

Now put all the given values in the above formula, we get the volume of mercury at $50^oC$ .

$V_{2}=0.002[1+0.00018(50-20)]=0.0020108m^3$

Therefore, the volume of mercury at $50^oC$ is, $0.002010812m^3$

7 0

3 years ago

When a diver gets into a tuck position by pulling in her arms and legs, she increases her angular speed. before she goes into th

Levart [38]

Before she gets into a tuck position, distribution of mass of her body about axis of rotation increases. As a result, her angular velocity decreases. When a diver gets into a tuck position, distribution of mass of her body about axis of rotation becomes less. As a result, her angular velocity increases. During this whole time, there is only one force acting on her body which is her own weight "mg". The net torque acting on her all the time is zero because "mg" force passes through this axis of rotation which causes no moment. Since there is no torque acting on her body, angular momentum should remain constant ( before tuck and after tuck)
Angular momentum is defined as,
L = Iω
Since net torque is zero,
Li = Lf ........... (Li = initial angular momentum, Lf = final angular momentum)
(Iω)i= (Iω)f

Ii = 2.0 kg · m² , ωi = 5.5 rad/s, ωf = <span>11.5 rad/s.
From given, find If. </span>

5 0

3 years ago